Question

Water from a tap emerges vertically down with an initial speed of $1.0{\text{ms}}^{-1}$. The cross-sectional area of tap is ${10}^{-4}{\text{m}}^2$. Assume that the pressure is constant throughout the stream of water, and that the flow is steady. The cross sectional area of the stream $0.15\text{m}$ below the tap is

• A. $5.0\times {10}^{-4}{\text{m}}^2$
• B. $1.0\times {10}^{-5}{\text{m}}^2$
• C. $5.0\times {10}^{-5}{\text{m}}^2$
• D. $2.0\times {10}^{-5}{\text{m}}^2$

Answer 1

The correct option is C $5.0\times {10}^{-5}{\text{m}}^2$

Applying bernoulli's theorem (energy conservation) between two points on the water stream,
$\text{Decrease in fluid's}PE=\text{increase in fluid's}KE$
$\Rightarrow \rho g\Delta h=\frac{1}{2}\rho (v_f^2-v_i^2)$
or $2\left(10\right)\left(0.15\right)=v_f^2-(1.0{)}^2$
$\because \Delta h=0.15\text{m}$, difference in height of two points is $0.15\text{m}$
$\Rightarrow v_f=2\text{m/s}$
Now from continuity equation
$A_1{v}_1=A_2{v}_2$
or $A\propto \frac{1}{v}$
velocity has become twice. Hence area of cross-section of water stream will become half.
$\Rightarrow A^{\prime }=\frac{A}{2}$
$\Rightarrow A^{\prime }=0.5\times {10}^{-4}{\text{m}}^2$
$\therefore A^{\prime }=5\times {10}^{-5}{\text{m}}^2$